\[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]

↓

\[\begin{array}{l} \mathbf{if}\;x \le -1.19339422339022175 \cdot 10^{-11}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}

↓

\begin{array}{l}
\mathbf{if}\;x \le -1.19339422339022175 \cdot 10^{-11}:\\
\;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\

\mathbf{else}:\\
\;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\

\end{array}

double code(double x) {
	return ((double) sqrt(((double) (((double) (((double) exp(((double) (2.0 * x)))) - 1.0)) / ((double) (((double) exp(x)) - 1.0))))));
}

↓

double code(double x) {
	double VAR;
	if ((x <= -1.1933942233902218e-11)) {
		VAR = ((double) sqrt(((double) (((double) (((double) (((double) exp(((double) (2.0 * x)))) - 1.0)) / ((double) fma(((double) -(1.0)), 1.0, ((double) exp(((double) (x + x)))))))) * ((double) (((double) exp(x)) + 1.0))))));
	} else {
		VAR = ((double) sqrt(((double) fma(0.5, ((double) pow(x, 2.0)), ((double) fma(1.0, x, 2.0))))));
	}
	return VAR;
}

Derivation

Split input into 2 regimes
if x < -1.1933942233902218e-11
1. Initial program 0.5
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Using strategy rm
3. Applied flip--0.3
  \[\leadsto \sqrt{\frac{e^{2 \cdot x} - 1}{\color{blue}{\frac{e^{x} \cdot e^{x} - 1 \cdot 1}{e^{x} + 1}}}}\]
4. Applied associate-/r/0.3
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{e^{x} \cdot e^{x} - 1 \cdot 1} \cdot \left(e^{x} + 1\right)}}\]
5. Simplified0.0
  \[\leadsto \sqrt{\color{blue}{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)}} \cdot \left(e^{x} + 1\right)}\]
if -1.1933942233902218e-11 < x
1. Initial program 36.1
  \[\sqrt{\frac{e^{2 \cdot x} - 1}{e^{x} - 1}}\]
2. Taylor expanded around 0 7.0
  \[\leadsto \sqrt{\color{blue}{0.5 \cdot {x}^{2} + \left(1 \cdot x + 2\right)}}\]
3. Simplified7.0
  \[\leadsto \sqrt{\color{blue}{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}}\]
Recombined 2 regimes into one program.
Final simplification0.8
\[\leadsto \begin{array}{l} \mathbf{if}\;x \le -1.19339422339022175 \cdot 10^{-11}:\\ \;\;\;\;\sqrt{\frac{e^{2 \cdot x} - 1}{\mathsf{fma}\left(-1, 1, e^{x + x}\right)} \cdot \left(e^{x} + 1\right)}\\ \mathbf{else}:\\ \;\;\;\;\sqrt{\mathsf{fma}\left(0.5, {x}^{2}, \mathsf{fma}\left(1, x, 2\right)\right)}\\ \end{array}\]

could not determine a ground truth for program body (more)

Error

Try it out

Derivation

`if x < -1.1933942233902218e-11`

`if -1.1933942233902218e-11 < x`

Reproduce

In
Out